
Note that the above proposal distribution \(\mathbf {z}\sim p_{\mathbf {z}}(\mathbf {z}), \mathbf {x}= G(\mathbf {z})\) is the same as \(p_{\mathbf {x}}(\mathbf {x})\)  by definition. Based on Lemma 1 in [1]} and Lemma 1 in [2]}, we arrive at:
\(p(\mathbf {x}|\mathcal {C}) &= \frac{p_{\mathbf {x}}(\mathbf {x})r(\mathbf {x})}{\mathbb {E}_{\mathbf {x}^{\prime } \sim p_{\mathbf {x}}(\mathbf {x}^{\prime })}[r(\mathbf {x}^{\prime })]} \\&= \frac{p_{\mathbf {x}}(\mathbf {x})e^{-E_{\mathcal {C}}(\mathbf {x})} / (M_{\mathcal {C}}\cdot Z)}{\mathbb {E}_{\mathbf {x}^{\prime } \sim p_{\mathbf {x}}(\mathbf {x}^{\prime })}[e^{-E_{\mathcal {C}}(\mathbf {x}^{\prime })} / (M_{\mathcal {C}}\cdot Z)]} \\&= \frac{p_{\mathbf {x}}(\mathbf {x})e^{-E_{\mathcal {C}}(\mathbf {x})}}{\mathbb {E}_{\mathbf {x}^{\prime } \sim p_{\mathbf {x}}(\mathbf {x}^{\prime })}[e^{-E_{\mathcal {C}}(\mathbf {x}^{\prime })}]} \\&= \frac{p_{\mathbf {x}}(\mathbf {x})e^{-E_{\mathcal {C}}(\mathbf {x})}}{Z_X}, \quad Z_X = \int _{\mathbf {x}^{\prime }} p_{\mathbf {x}}(\mathbf {x}^{\prime })e^{-E_{\mathcal {C}}(\mathbf {x}^{\prime })} d\mathbf {x}^{\prime }.\) 
